Section 5Chapter 6

1 Copyright © 2012, 2008, 2004 Pearson Education, Inc. Objectives 2 3 Solving Equations by Factoring Learn and use the zero-factor property. Solve applied problems that require the zero-factor property. Solve a formula for a specified variable, where factoring is necessary. 6.5

Copyright © 2012, 2008, 2004 Pearson Education, Inc. Solve the equation. 3x 2 – x = 4 Step 1 Standard form. 3x 2 – x – 4 = 0 Step 2 Factor.(3x – 4)(x + 1) = 0 Step 3 Zero-factor.3x – 4 = 0 or x + 1 = 0 Step 4 Solve. Step 5 Check.3x 2 – x = 4 3x 2 – x = 4 Slide CLASSROOM EXAMPLE 1 Solving Quadratic Equations by Factoring Solution: The solution set is True

Copyright © 2012, 2008, 2004 Pearson Education, Inc. Double Root – A quadratic equation that has the same solution twice. Objective 1 Slide

Copyright © 2012, 2008, 2004 Pearson Education, Inc. Solve the equation. 16m m + 9 = 0 ** It’s a perfect square trinomial ** Step 1 Standard form. Already in standard form. Step 2 Factor. (4m + 3) 2 = 0 Step 3 Zero-factor. 4m + 3 = 0 4m +3 = 0 Step 4 Solve. 4m = -3 twice double root** Step 5 Check. The solution set is Slide CLASSROOM EXAMPLE 2 Solving Quadratic Equations by Factoring (cont’d) Solution: True

Copyright © 2012, 2008, 2004 Pearson Education, Inc. Solve x x = 0 Factor.x(x + 12) = 0 x = 0 or x + 12 = 0 x = 0 or x = –12 Check. x x = 0 x x = (0) = 0 (−12) (–12) = 0 0 = 0 0 = 0 The solution set is {–12, 0}. Slide CLASSROOM EXAMPLE 3 Solving a Quadratic Equation with a Missing Constant Term Solution: True

Copyright © 2012, 2008, 2004 Pearson Education, Inc. Solve 5x 2 – 80 = 0 5x 2 – 80 = 5(x 2 – 16) Factor out 5. = 5(x – 4)(x + 4) Factor. x – 4 = 0 or x + 4 = 0Solve. x = 4 or x = – 4 Check that the solution set is {– 4, 4}. Slide CLASSROOM EXAMPLE 4 Solving a Quadratic Equation with a Missing Linear Term Solution:

Copyright © 2012, 2008, 2004 Pearson Education, Inc. Solve 3x 3 + x 2 = 4x 3x 3 + x 2 – 4x = 0Standard form. x(3x 2 + x – 4) = 0 Factor out n. x(3x + 4)(x – 1) = 0 x = 0 or 3x + 4 = 0 or x – 1 = 0 x = 1 Check that the solution set is Slide CLASSROOM EXAMPLE 6 Solving an Equation of Degree 3 Solution:

Copyright © 2012, 2008, 2004 Pearson Education, Inc. Solve applied problems that require the zero-factor property. Objective 2 Slide

Copyright © 2012, 2008, 2004 Pearson Education, Inc. The length of a room is 2 m less than three times the width. The area of the room is 96 m 2. Find the width of the room. Step 1 Read the problem again. There will be one answer. Step 2 Assign a variable. Let x = the width of the room. Let 3x – 2 = the length of the room. The area is 96. x 3x – 2 Slide CLASSROOM EXAMPLE 7 Using a Quadratic Equation in an Application Solution:

Copyright © 2012, 2008, 2004 Pearson Education, Inc. Step 3 Write an equation. The area of a rectangle is the length times the width. A = lw 96 = (3x – 2)x Step 4 Solve. 96 = (3x – 2)x 96 = 3x 2 – 2x 0 = 3x 2 – 2x – 96 0 = (3x + 16)(x – 6) 3x + 16 = 0 or x – 6 = 0 x = 6 Slide CLASSROOM EXAMPLE 7 Using a Quadratic Equation in an Application (cont’d) x 3x – 2

Copyright © 2012, 2008, 2004 Pearson Education, Inc. Step 5 State the answer. A distance cannot be negative, so reject as a solution. The only possible solution is 6, so the width of the room is 6 m. Step 6 Check. The length is 2 m less than three times the width, the length would be 3(6) – 2 = 16. The area is 6 16 = 96, as required. Slide CLASSROOM EXAMPLE 7 Using a Quadratic Equation in an Application (cont’d) x 3x – 2 When application lead to quadratic equations, a solution of the equation may not satisfy the physical requirements of the problem. Reject such solutions as valid answers.

Copyright © 2012, 2008, 2004 Pearson Education, Inc. If a small rocket is launched vertically upward from ground level with an initial velocity of 128 ft per sec, then its height in feet after t seconds is a function defined by h(t) = –16t t if air resistance is neglected. How long will it take the rocket to reach a height of 256 feet? We let h(t) = 256 and solve for t. 256 = –16t t 16t 2 – 128t = 0 Divide by 16. t 2 – 8t + 16 = 0 (t – 4) 2 = 0 t – 4 = 0 t = 4 It will reach a height of 256 feet in 4 seconds. Slide CLASSROOM EXAMPLE 8 Using a Quadratic Function in an Application Solution:

Copyright © 2012, 2008, 2004 Pearson Education, Inc. Solve a formula for a specified variable, where factoring is necessary. Objective 3 Slide

Copyright © 2012, 2008, 2004 Pearson Education, Inc. Solve the formula for W. Slide CLASSROOM EXAMPLE 9 Using Factoring to Solve for a Specified Variable Solution: We must write the expression so that the specified variable is a factor. Then we can divide by its coefficient in the final step.